Question

One half mole of a monoatomic ideal gas absorbs 1200 j of heat energy while performing 2196 j of work. The temperature of the gas changes by

Answer 1

Answer:

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Answer 2

Answer:

The temperature change of the given monoatomic ideal gas, ΔT, calculated is $-160K$.

Explanation:

Data given,

The heat absorbed by the monoatomic ideal gas, ΔH = $1200J$

The work done on the system by a monoatomic ideal gas, ΔW = $2196J$

The number of moles of monoatomic gas, n = $\frac{1}{2} mol$

The temperature change of the given monoatomic ideal gas, ΔT =?

Now, from the relation given below, we can find out the enthalpy:

• $\Delta H= \Delta U + \Delta W$
• $1200 = U + 2196$
• ΔU = $-996J$

Now, we know that;

• ΔU = $nC_{V} \Delta T$

Here, $C_{V}$ = heat capacity at constant volume

And $C_{V}$ for monoatomic gas = $\frac{3}{2} R$     ( R = Raydberg's constant = $8.314 J/K-mol$ )

Therefore, the equation becomes:

• ΔU = $n\times \frac{3}{2}R\times \Delta T$
• $-996 = \frac{1}{2} \times \frac{3}{2}\times 8.314\times \Delta T$
• $-996 = \frac{24.9}{4}\times \Delta T$
• $\Delta T = -996\times \frac{4}{24.942}$ = $-160K$

Hence, the temperature change of the given monoatomic ideal gas, ΔT = $-160K$.